I'm studying for finals at the moment and could use some help with solving the particular solution for this system of nonhomogeneous differential equations:

$x' = \begin{bmatrix}1 & 0\\ 2 & -3 \end{bmatrix}x + \begin{bmatrix}10\cos(t)\\ 0\end{bmatrix}$

Can someone walk me through what needs to be done? I know that the particular solution will take the form: $x_p(t) = A\cos(t) + B\sin(t)$, but apart from that, I'm unsure of how to proceed.


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    $\begingroup$ I see you have accepted some answers, so let me help you out. Do you know how to compute matrix exponentials? It's not necessary to solve this, there is a very simple way, but computing matrix exponentials allows you to attack any kind of problem like this. $\endgroup$ – Git Gud May 1 '14 at 0:21
  • $\begingroup$ I'm not familiar with that terminology, sorry. I'm trying to go about this using the method of underdetermined coefficients. $\endgroup$ – TheColonel May 1 '14 at 0:22
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    $\begingroup$ Do you understand what's going on in this answer? $\endgroup$ – Git Gud May 1 '14 at 0:26
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    $\begingroup$ @Git Gud, I arrived at this solution for the particular solution: x_p_(t) = {{-5},{-4}}cos(t) + {{5},{2}}sin(t) Is this correct? $\endgroup$ – TheColonel May 1 '14 at 0:33
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    $\begingroup$ Your particular solution is correct. $\endgroup$ – Git Gud May 1 '14 at 0:58

We have:

$$\phi(t) = \left( \begin{array}{cc} 0 & 2 e^t \\ e^{-3 t} & e^t \\ \end{array} \right)$$

$$\phi^{-1}(t) = \left( \begin{array}{cc} -\frac{e^{3 t}}{2} & e^{3 t} \\ \frac{e^{-t}}{2} & 0 \\ \end{array} \right)$$

$$\phi^{-1}(t).\left( \begin{array}{c} 10 \cos (t) \\ 0 \\ \end{array} \right) = \left( \begin{array}{c} -5 e^{3 t} \cos (t) \\ 5 e^{-t} \cos (t) \\ \end{array} \right)$$

Integrating the previous result yields:

$$\left( \begin{array}{c} -\frac{1}{2} e^{3 t} (3 \cos (t)+\sin (t)) \\ \frac{5}{2} e^{-t} (\sin (t)-\cos (t)) \\ \end{array} \right)$$

So we have:

$$x_p(t) = \phi(t).\left( \begin{array}{c} -\frac{1}{2} e^{3 t} (3 \cos (t)+\sin (t)) \\ \frac{5}{2} e^{-t} (\sin (t)-\cos (t)) \\ \end{array} \right) = \left( \begin{array}{c} 5 (\sin (t)-\cos (t)) \\ 2 \sin (t)-4 \cos (t) \\ \end{array} \right)$$


$$x(t) = x_h(t) + x_p(t)$$

This yields:

$$x(t) = c_1 e^t+5 \sin (t)-5 \cos (t) \\ y(t) = \frac{1}{2} \left(e^{-3 t} \left(c_1 \left(e^{4 t}-1\right)+2 c_2\right)+4 \sin (t)-8 \cos (t)\right)$$

  • $\begingroup$ Mathematica is a good thing, especially when you copy what it produces without typos. $\endgroup$ – Artem May 2 '14 at 13:39

Set $x=\begin{bmatrix} x_1 \\ x_2\end{bmatrix}, A=\begin{bmatrix} 1 & 0 \\ 2 & -3\end{bmatrix}$ and $b(t)=\begin{bmatrix} 10\cos (t) \\ 0\end{bmatrix}$, for all $t\in \mathbb R$.

Your system is equivalent to $x'=Ax+b$.

Given $t\in \mathbb R$, it can be rewritten as $$\begin{cases} x_1'(t)&=x_1(t)+ 10\cos (t)\\ x_2'(t)&=2x_1(t)-3x_2(t)\end{cases}$$

The equation $x_1'(t)=x_1(t)+ 10\cos (t)$ is just an ODE. It can be solved using the standard methods, (for instance this one). The general solution of this equation is determined by $$\forall t\in \mathbb R\left(x_1(t)=5\sin(t)-5\cos(t) +C_1e^t\right),$$ for some $C_1\in \mathbb R$.

So now you can just replace $x_1(t)$ by the above expression in the second equation yielding:


This, again, is just an ODE. The solution is given by, for some $C_2\in \mathbb R$, $$\forall t\in \mathbb R\left(x_2(t)=2\sin(t)-4\cos(t)+\dfrac{C_1}{2}e^t+C_2e^{-3t}\right).$$

A particular solution to the system can be found setting $C_1=C_2=0$ and this is exactly the solution you found.

Since you're looking to solve this with the indeterminate coefficients method, let me also do this.

Take the ansatz $x_p=B\cos +C\sin $, where $B,C\in \mathbb R^{2\times 1}$ are constants.
Write $B=\begin{bmatrix} b_1\\ b_2\end{bmatrix}, C=\begin{bmatrix} c_1\\ c_2\end{bmatrix}$ and $x_p(t)=\begin{bmatrix} b_1\cos(t)+c_1\sin (t)\\ b_2\cos(t)+c_2\sin(t)\end{bmatrix}$

For all $t\in \mathbb R$ set $$\begin{bmatrix} -b_1\sin(t)+c_1\cos (t)\\ -b_2\sin(t)+c_2\cos(t)\end{bmatrix}=x_p'(t)=Ax_p(t)+b(t)=\begin{bmatrix} b_1\cos(t) + c_1\sin(t)+10\cos(t)\\ (2b_1-3b_2)\cos(t)+(2c_1-3c_2)\sin(t)\end{bmatrix},$$

which is equivalent to $$\begin{cases} (10+b_1-c_1)\cos(t)+(c_1+b_1)\sin(t)&=0\\ (2b_1-3b_2-c_2)\cos(t)+(2c_1-3c_2+b_2)\sin(t)&=0\end{cases} \tag {$\spadesuit$}$$

Since $\left(\spadesuit\right)$ holds for all $t$, it holds in particular for $t=0$ and $t=\dfrac\pi 2$ thus providing the system $$\begin{cases} b_1+ c_1&=0\\ b_1-c_1&=-10\\ 2b_1-3b_2-c_2&=0\\ b_2+2c_1-3c_2&=0\end{cases}$$

whose solution is attained with $(b_1, b_2, c_1, c_2)=(-5, -4, 5, 2)$.

Therefore $x_p(t)=\begin{bmatrix} -5\cos(t)+5\sin (t)\\ -4\cos(t)+2\sin(t)\end{bmatrix},$ for all $t\in \mathbb R$.


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